# Lorentz transforms

1. Aug 12, 2007

### ehrenfest

Zwiebach page 36

a)Give the Lorentz transformations for the components a_mu of a vector under a boost along the x^1 axis.

$$a'_0 = -a_0 \gamma + a_1 \gamma \beta$$
$$a'_1^ = -a_0 \gamma \beta + a_1 \gamma$$
and 2 and 3 are the same.

b) Show that the objects

$$\partial/\partial x^\mu$$ transform under a boost along the x^1 axis in the same way as the $$a_\mu$$ in (a) do

Note the upper and lower indices.

Firstly is my answer to (a) good? Second, I am not sure how to transform a partial derivative in (b).

2. Aug 13, 2007

### George Jones

Staff Emeritus
I get a different result. Careful with th signs.

Use the multivariable chain rule.

3. Aug 13, 2007

### ehrenfest

Is this better for part (a)
$$a'_0 = a_0 \gamma + a_1 \gamma \beta$$
$$a'_1 = a_0 \gamma \beta + a_1 \gamma$$

?

4. Aug 14, 2007

### George Jones

Staff Emeritus
Yes, this is what I get.

5. Aug 17, 2007

### ehrenfest

I get, for the example of mu = 0:

$$\frac{\partial}{\partial x'^0} = \frac{\partial}{\partial x'^0} \frac{\partial x' ^ 0}{\partial x^0} + \frac{\partial}{\partial x'^0} \frac{\partial x' ^ 0}{\partial x^1}$$

How do I get partials with respect to non-primed coordinates with the multivariable chain rule?

6. Aug 17, 2007

### George Jones

Staff Emeritus
Do you know a formula that expresses $x'^0$ in terms of $x^0$ and $x^1?$ If you do, then just differentiate.

7. Aug 18, 2007

### ehrenfest

Like this:

$$\frac{\partial x'^0}{\partial x^0} = \gamma$$

$$\frac{\partial x'^0}{\partial x^1} = -\gamma \beta$$

In order to show that the objects

$$\partial/\partial x^\mu$$

transform as suggested, I need to express the partial operators with respect to the primed coordinates in terms of the partial operators with respect to the nonprimed coordinates, right?

What I showed above does not even have an operator really.

8. Aug 20, 2007

### George Jones

Staff Emeritus
Substitute these into a corrected version of the equation that you gave in post #5. I just noticed that this equation is not right.

9. Aug 23, 2007

### ehrenfest

I honestly do not see why my multivariable chain rule equation is wrong.

10. Aug 23, 2007

### George Jones

Staff Emeritus
In post #5, you wrote

"Cancellation" (I don't really mean cancellation) should turn each term on the right into the term on the left. After "cancellation' your expression is

$$\frac{\partial}{\partial x'^0} = \frac{\partial}{\partial x^0} + \frac{\partial }{\partial x^1}.$$

Notice that each term on the right looks different than the term on the left.

Take a close look at the chain rule in a text.

11. Aug 24, 2007

### ehrenfest

I see. I am good with part (b). Now, for part (c), he means the expressions

p = -i h-bar del

and E = i h-bar d/dt

right?

So the p_mu is the momentum four-vector? How can you have a momentum four-vector when one component of that vector is basically time?

12. Aug 28, 2007

### George Jones

Staff Emeritus
With lower indices. Forget about quantum theory for a bit. In an inertial frame, what is p_0?

13. Aug 28, 2007

### ehrenfest

$$p_0 = -E/c$$

So I need to show that this is also equal to $$\hbar/i \frac{\partial}{\partial x^{\mu}}$$? Should I use those equation in my previous post? Plugging in for E does not seem to work.

Last edited: Aug 28, 2007
14. Aug 29, 2007

### George Jones

Staff Emeritus
You need to show that it is equal to

$$\hbar/i \frac{\partial}{\partial x^0}$$

Yes.

15. Aug 29, 2007

### ehrenfest

Got it. Done with 2.3!

16. Aug 30, 2007

### George Jones

Staff Emeritus
Great!

Note that Zwiebach uses "physicist's math", i.e., he treats each $\partial / \partial x^\mu$ as a component of a covector. In modern (not so modern now) differential geometry, each $\partial / \partial x^\mu$ is a tangent vector, not a component.